Motion in 2-D

• Projectiles

• Circular Motion

• Updated 10.12.2014

Projectile Motion - what do you notice?

Two kinds of motion blended together

• Vertical - acceleration

• Horizontal - constant velocity

The Return of the Big 5 (in part)

• Vertical

• d = vit + ½ gt 2

• vf = vi + gt

• vf = vi2 +2gd

• Horizontal• V = d/t

How long will it take a ball to fall when thrown at 20. m/s horizontally

from a height of 15 m? How far horizontally will it travel?

• Draw the situation and label with given information

• Decide upon formulae to use

• Solve using formulae and showing UNITS!

Time to fall

d = vit + ½ gt 2

Solve for t

Initial v in the vertical is 0 so

t = sr (2d/g)

T = sr (2(15m)/9.8m/s2)

T = 1.7 s

How far horizontally will it travel?

d = vt

d = 20m/s (1.7s)

d = 34 m

A ball is launched at 4.5 m/s at 66° above the horizontal.What are the maximum height and flight time of the ball?

Establish a coordinate system with the initial position of the ball at the origin.

Show the positions of the ball at the beginning, at the maximum height, and at the end of the flight.

• Known Unknowndy init = 0 h max = dy = ?

= 66 o total flight time, t = ?

vinit= 4.5 m/s

a y = - g

• Known Unknowndy init = 0 h max = dy = ?

= 66 o total flight time, t = ?

vinit= 4.5 m/s

a y = - g

Find Vy using sin:

Sin = opp/hyp = Vy/4.5m/s

Vy = 4.5m/s sin 66o

Vy = 4.1m/s

Vyf = Vi + gt Vyf = 0 (it stops)

t = -Vi/-g = 4.1m/s/9.8m/s2 =

t = 0.42 s

Double this for up and down motion to get so 2 x 0.42 s

total flight time = 0.84 s

Find the y-component of vi.

v yi = v i (sin )

= 4.5 m/s (sin 66)

= 4.1 m/s

dy = ?

dy = ½ g t2 = ½ (9.8 m/s2)(0.42s)2

dy = 0.86 m = 8.6 x 10-1 m

Ch 6: 2D Motion Assignment

• Projectiles: 1, 2, 3, 4, 51-53

Projectile Simulation

http://www3.interscience.wiley.com:8100/legacy/college/halliday/0471320005/simulations6e/index.htm

Circular MotionAn object moving in a circle at a constant speed is accelerated

Centripetal acceleration depends upon the object’s speed and the radius of the circle

Centripetal force causes centripetal acceleration.

Ch 6 Circular Motion Problems: 12-20, 49,61-63

V

acFc

Definitions

• Centripetal Force - a center seeking force that appears to pull an object toward the center of the circle along which it is moving

• Centripetal Acceleration - acceleration toward the center ac = v2/r

Newton’s Law enters in…

• Since f = ma

• The centripetal force is determined by substituting

• ac = v2/r a = f/m

• fc = mv2/r